Question

A moving block having mass $m$ collides with another stationary block having mass $4m$. The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v$, then the value of coefficient of restitution (e) will be

• A. $0.5$
• B. $0.25$
• C. $0.4$
• D. $0.8$

Answer 1

The correct option is B

$0.25$

Step 1: Given

Mass of the first block: $m_1=m$

Mass of the second block: $m_2=4m$

The initial velocity of the first block: $u_1=v$

The initial velocity of the second block: $u_2=0$

The final velocity of the first block: $v_1=0$

Let the final velocity of the second block: $v_2=v\text{'}$

Step 2: Formula Used

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

$e=\frac{Velocityofseparation}{Velocityofapproach}$

Step 3: Calculate the final velocity of the second block

Substitute the given values in the linear law of conservation of momentum

$$\begin{gathered} m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2 \\ \Rightarrow mv+4m\times 0=m\times 0+4mv\text{'} \\ \Rightarrow mv=4mv\text{'} \\ \Rightarrow v\text{'}=\frac{v}{4} \end{gathered}$$

Step 4: Calculate the coefficient of restitution

Substitute v for the velocity of approach and v' for the velocity of separation.

$$\begin{gathered} e=\frac{Velocityofseparation}{Velocityofapproach} \\ =\frac{v\text{'}}{v} \\ =\frac{{\displaystyle \frac{v}{4}}}{v} \\ =\frac{{\displaystyle 1}}{4} \\ =0.25 \end{gathered}$$

Hence, the value of the coefficient of restitution is $0.25$.

Thus, option (B) is correct.